cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A365093 Smallest k such that A365092(k) = n.

Original entry on oeis.org

2, 3, 4, 5, 7, 10, 11, 20, 22, 23, 43, 46, 47, 92, 94, 139, 188, 235, 283, 461, 517, 659, 941, 1081, 1319, 2027, 2447, 2879, 4139, 5758, 8278, 10343, 13301, 20117, 26179, 30337, 44227, 56281, 61993, 95197, 115009, 135313, 194533, 270626, 366683, 481199, 606743, 811879
Offset: 2

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Author

Jianing Song, Aug 21 2023

Keywords

Comments

a(n) is well-defined for all n >= 2.
Sequence is increasing: write a(n) as the sums and products of 1s by the process described in A365092. Removing any 1 in the expression yields a smaller number.
For m,n >= 2, it is easy to see that if gcd(a(m),a(n)) = 1, then a(m+n) <= a(m)*a(n). This is conjectured to be true for all m,n. In particular, it is conjectured that a(n+2) <= 2*a(n). If a(m+n) <= a(m)*a(n) is true, then by Fekete's subadditive lemma, we have lim_{n->oo} a(n)^(1/n) = inf_{n>=1} a(n)^(1/n) <= a(29)^(1/29) = 2879^(1/29) = 1.3160857758...
Are all terms other than 4, 20, 92 and 188 squarefree?
What are the primes that divide infinitely many terms? In particular, for p = 2 or 47, are there infinitely many terms divisible by p? Is there any term divisible by 3, 5, 7, 11 or 43 other than themselves, 235 and 517?

Examples

			a(9) = 20 since 20 = (1+1)^(1+1)*((1+1)^(1+1)+1) (see A365092) uses nine 1s, and all smaller numbers use fewer than nine 1s.
		

Crossrefs

Cf. A365092.

Programs

  • PARI
    A365093_search(lim) = my(list=[], k=2); for(n=1, oo, if(A365092(n)==k, list=concat(list,n); k++); if(k==lim, return(list))) \\ see A365092 for its program
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